An air conditioner removes heat steadily from a house at a rate of 700 kJ/min while drawing electric power at a rate of 5.10 kW. Determine: a. the COP of this air conditioner
b. the rate of heat transfer to the outside air.

a) The Coefficient of Performance of this air conditioner is 2.288.

b) The rate of heat transfer to the outside air is 16.767 kilowatts (1006.02 kilojoules per minute).

Explanation:

a) Air conditioners are applications of refrigeration cycles, whose performance is analyzed by means of the Coefficient of Performance (), dimensionless:

(Eq. 1)

Where:

- Heat removal rate from the house, measured in kilowatts.

- Electric power, measured in kilowatts.

If we know that and , the Coefficient of Performance of this air conditioner is:

The Coefficient of Performance of this air conditioner is 2.288.

b) We find the rate of heat transfer to the outside air (), measured in kilowatts, by applying the First Law of Thermodynamics:

(Eq. 2)

If we get that and , then the rate of heat transfer to the outside air is:

The rate of heat transfer to the outside air is 16.767 kilowatts (1006.02 kilojoules per minute).

answer:

the formula for amps is watts divided by volts. to use the chart, cover up the a with your finger and use the remaining chart calculation of w divided by v. using our sample panel data, 60 watts divided by 12 volts equals 5 amps.

explanation: